# Issue with integral in dimensional regularization

• Karatechop250
In summary, the conversation discusses a specific integral that arises in calculating renormalization group equations for exotic particles. The integral cannot be solved using dimensional regularization, so alternative methods are explored. One approach is to introduce an intermediate scale and separate the integral into terms with IR and UV divergences. These divergences can be dealt with using appropriate counterterms and renormalization schemes.
Karatechop250
So I am calculating the renormalization group rquations for some exotic new particles and used dimensional regularization for all the calculations up to this point. Now I am look at the vertex corrections in the massless limit in which all external momentum are equal to 0. The issue here is I end up with an integral I don't know how to solve.

Its this integral
$$\int d^{4}l \frac{1}{l^4}$$
which you obviously can't solve using dimensional regularization. What other method can I use to solve it and how will I make the solution compatable with my other solutions since they will have the ##\frac{1}{\epsilon}##

Well, I don't know much about renormalization, but that integral should be no different than ##\int \frac{1}{\ell^4}dl*dl*dl*dl##
I drew that from the ##d\tau## notation that means ##\int d\tau = \int\int\int dx*dy*dz## Does that help you solve it?

Karatechop250 said:
So I am calculating the renormalization group rquations for some exotic new particles and used dimensional regularization for all the calculations up to this point. Now I am look at the vertex corrections in the massless limit in which all external momentum are equal to 0. The issue here is I end up with an integral I don't know how to solve.

Its this integral
$$\int d^{4}l \frac{1}{l^4}$$
which you obviously can't solve using dimensional regularization. What other method can I use to solve it and how will I make the solution compatable with my other solutions since they will have the ##\frac{1}{\epsilon}##

What does that integral mean? Usually when people write something like $\int dx^3$, that's shorthand for $\int dx \int dy \int dz$. I don't know what you mean by $\int d^4 l$

BiGyElLoWhAt
stevendaryl said:
What does that integral mean? Usually when people write something like $\int dx^3$, that's shorthand for $\int dx \int dy \int dz$. I don't know what you mean by $\int d^4 l$

Hmm. Maybe I'm forgetting the convention. It looks like from Bill's post that $\int d^3 x$ is used, rather than $\int dx^3$. But in either case, it's not clear what $\int d^4 l \frac{1}{l^4}$ means. Is this supposed to be the same as $\int dl1 \int dl2 \int dl3 \int dl4 \frac{1}{\sqrt{l1^2 + l2^2 + l3^2 + l4^2}}$?

The issue is the metric defining l^2.

It some instances, and this is one, its clearer putting the dx etc in front.

Thanks
Bill

Could someone possibly shed some context on what's happening here? So l is in a curved space? I don't see how putting the differential(s) in front of the function clears things up. I'm apparently missing the point.

BiGyElLoWhAt said:
Could someone possibly shed some context on what's happening here? So l is in a curved space? I don't see how putting the differential(s) in front of the function clears things up. I'm apparently missing the point.

Thanks
Bill

bhobba said:

I skimmed it, and it doesn't provide context. Are you referring to the F(k^2) generalization? It goes into details on how to solve it, but says nothing about the context. Thanks anyways.

BiGyElLoWhAt said:
I skimmed it, and it doesn't provide context. Are you referring to the F(k^2) generalization? It goes into details on how to solve it, but says nothing about the context.

They are integrals like you get in a math book. That's its context.

Thanks
Bill

Karatechop250 said:
So I am calculating the renormalization group rquations for some exotic new particles and used dimensional regularization for all the calculations up to this point. Now I am look at the vertex corrections in the massless limit in which all external momentum are equal to 0. The issue here is I end up with an integral I don't know how to solve.

Its this integral
$$\int d^{4}l \frac{1}{l^4}$$
which you obviously can't solve using dimensional regularization. What other method can I use to solve it and how will I make the solution compatable with my other solutions since they will have the ##\frac{1}{\epsilon}##

You can treat this in dimensional regularization. Typically you would just say that this dimensionless integral gives zero by the Veltman rule, but we can be more specific. The interesting part is obviously

$$I = \int_0^\infty d\ell \ell^{d-5} = \left[ \frac{k^{d-4}}{d-4} \right]_0^\infty,$$

which is UV divergent if ##d>4## and IR divergent.if ##d<4##. We can isolate the divergences by introducing an intermediate scale ##\mu## and writing the integral as the sum

$$I = \int_0^\mu d\ell \ell^{d-5} + \int_\mu^\infty d\ell \ell^{d-5}.$$

The first term contains the IR divergence so we let ##d = 4 + \epsilon_\text{IR}##, with ##\epsilon_\text{IR}>0##. The 2nd term has the UV divergence, so there we let ##d=4-\epsilon_{UV}##. Putting these together, we find

$$I = \int_0^\mu d\ell \ell^{\epsilon_\text{IR}-1} + \int_\mu^\infty d\ell \ell^{-\epsilon_{UV}-1} = \frac{\mu^{\epsilon_\text{IR}}}{\epsilon_\text{IR}} + \frac{1}{\epsilon_\text{UV} \mu^{\epsilon_\text{UV}}} .$$

For small ##\epsilon_\text{IR},\epsilon_\text{UV}##,

$$I \approx \frac{1}{\epsilon_\text{IR}} + \log \mu + \frac{1}{\epsilon_\text{UV} } - \log \mu \approx \frac{1}{\epsilon_\text{IR}} + \frac{1}{\epsilon_\text{UV} }.$$

How to deal with these divergences now depends on the physical process under consideration. The UV divergence should be automatically canceled by the appropriate counterterm for the diagram being considered. For the IR divergence, we can simply include the subtraction in the renormalization scheme, e.g. MS. Otherwise we have to incorporate the diagram into a physical quantity. Then the possible IR divergences will cancel in the sum over the collection of diagrams that contribute to the physical quantity.

## What is dimensional regularization?

Dimensional regularization is a mathematical technique used to deal with divergent integrals in quantum field theory. It involves analytically continuing the number of spacetime dimensions from the usual 4 dimensions to a complex number, allowing for the cancellation of infinities.

## What is the issue with integrals in dimensional regularization?

The issue with integrals in dimensional regularization is that the resulting integrals may still have divergences in certain cases, particularly in higher-order calculations. This can lead to inconsistencies in the final calculations and require additional techniques to fully resolve.

## How does dimensional regularization differ from other regularization methods?

Dimensional regularization differs from other regularization methods in that it is based on analytically continuing the number of spacetime dimensions, rather than introducing a cutoff or modifying the underlying theory. This makes it a more elegant and versatile technique, but also potentially more complex to implement.

## What are the benefits of using dimensional regularization?

The benefits of using dimensional regularization include its ability to handle a wide range of divergences, its compatibility with gauge theories, and its preservation of important symmetries such as Lorentz invariance. It also leads to simpler and more elegant calculations compared to other regularization methods.

## What are some limitations of dimensional regularization?

One limitation of dimensional regularization is that it may not always result in a physically meaningful answer, particularly when dealing with non-renormalizable theories. It also requires careful handling of poles and branch cuts in the analytic continuation process. Additionally, it may not be the most efficient method for calculations involving large numbers of loops.

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